3.1.43 \(\int \sin (2 x) \sin (4 x) \, dx\) [43]

Optimal. Leaf size=17 \[ \frac {1}{4} \sin (2 x)-\frac {1}{12} \sin (6 x) \]

[Out]

1/4*sin(2*x)-1/12*sin(6*x)

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Rubi [A]
time = 0.01, antiderivative size = 17, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, integrand size = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {4367} \begin {gather*} \frac {1}{4} \sin (2 x)-\frac {1}{12} \sin (6 x) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sin[2*x]*Sin[4*x],x]

[Out]

Sin[2*x]/4 - Sin[6*x]/12

Rule 4367

Int[sin[(a_.) + (b_.)*(x_)]*sin[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[Sin[a - c + (b - d)*x]/(2*(b - d)), x]
- Simp[Sin[a + c + (b + d)*x]/(2*(b + d)), x] /; FreeQ[{a, b, c, d}, x] && NeQ[b^2 - d^2, 0]

Rubi steps

\begin {align*} \int \sin (2 x) \sin (4 x) \, dx &=\frac {1}{4} \sin (2 x)-\frac {1}{12} \sin (6 x)\\ \end {align*}

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Mathematica [A]
time = 0.01, size = 17, normalized size = 1.00 \begin {gather*} \frac {1}{4} \sin (2 x)-\frac {1}{12} \sin (6 x) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sin[2*x]*Sin[4*x],x]

[Out]

Sin[2*x]/4 - Sin[6*x]/12

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Maple [A]
time = 0.03, size = 9, normalized size = 0.53

method	result	size
derivativedivides	\(\frac {\left (\sin ^{3}\left (2 x \right )\right )}{3}\)	\(9\)
default	\(\frac {\left (\sin ^{3}\left (2 x \right )\right )}{3}\)	\(9\)
risch	\(\frac {\sin \left (2 x \right )}{4}-\frac {\sin \left (6 x \right )}{12}\)	\(14\)
norman	\(\frac {\frac {2 \tan \left (x \right ) \left (\tan ^{2}\left (2 x \right )\right )}{3}-\frac {\left (\tan ^{2}\left (x \right )\right ) \tan \left (2 x \right )}{3}-\frac {2 \tan \left (x \right )}{3}+\frac {\tan \left (2 x \right )}{3}}{\left (1+\tan ^{2}\left (x \right )\right ) \left (1+\tan ^{2}\left (2 x \right )\right )}\)	\(51\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(2*x)*sin(4*x),x,method=_RETURNVERBOSE)

[Out]

1/3*sin(2*x)^3

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Maxima [A]
time = 1.07, size = 13, normalized size = 0.76 \begin {gather*} -\frac {1}{12} \, \sin \left (6 \, x\right ) + \frac {1}{4} \, \sin \left (2 \, x\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(2*x)*sin(4*x),x, algorithm="maxima")

[Out]

-1/12*sin(6*x) + 1/4*sin(2*x)

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Fricas [A]
time = 0.65, size = 14, normalized size = 0.82 \begin {gather*} -\frac {1}{3} \, {\left (\cos \left (2 \, x\right )^{2} - 1\right )} \sin \left (2 \, x\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(2*x)*sin(4*x),x, algorithm="fricas")

[Out]

-1/3*(cos(2*x)^2 - 1)*sin(2*x)

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Sympy [A]
time = 0.12, size = 22, normalized size = 1.29 \begin {gather*} - \frac {\sin {\left (2 x \right )} \cos {\left (4 x \right )}}{3} + \frac {\sin {\left (4 x \right )} \cos {\left (2 x \right )}}{6} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(2*x)*sin(4*x),x)

[Out]

-sin(2*x)*cos(4*x)/3 + sin(4*x)*cos(2*x)/6

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Giac [A]
time = 0.71, size = 8, normalized size = 0.47 \begin {gather*} \frac {1}{3} \, \sin \left (2 \, x\right )^{3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(2*x)*sin(4*x),x, algorithm="giac")

[Out]

1/3*sin(2*x)^3

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Mupad [B]
time = 0.17, size = 13, normalized size = 0.76 \begin {gather*} \frac {\sin \left (2\,x\right )}{4}-\frac {\sin \left (6\,x\right )}{12} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(2*x)*sin(4*x),x)

[Out]

sin(2*x)/4 - sin(6*x)/12

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